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3 years ago

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The following molecule can be synthesized from an epoxide and alkyllithium reagent, followed by aqueous workup. However, there a

re two ways in which this molecule can be formed. In one approach, the epoxide has a higher molecular weight than the epoxide used in the other approach. Draw both versions in the boxes below. Draw the carbon-lithium bond as a covalent bond.Part 1:Draw the combination of epoxide and alkyllithium reagent that uses a lower molecular weight epoxide.Part 2: Draw the combination of epoxide and alkyllithium reagent that uses a higher molecular weight epoxide.

1 answer:

Assoli18 [71]3 years ago

8 0

X. X

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I. I
I I. I.

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A transverse wave on a long horizontal rope with a

frosja888 [35]

Answer:

2 seconds

Explanation:

The frequency of a wave is related to its wavelength and speed by the equation

$f=\frac{v}{\lambda}$

where

f is the frequency

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For the wave in this problem,

v = 2 m/s

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So the frequency is

$f=\frac{2}{8}=0.25 Hz$

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$T=\frac{1}{f}=\frac{1}{0.25}=4 s$

This means that the wave takes 4 seconds to complete one full cycle.

Therefore, the time taken for the wave to go from a point with displacement +A to a point with displacement -A is half the period, therefore for this wave:

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2 years ago

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Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :

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Here, $I_2=I_1+mr^2$

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Solving the above equation, we get the value as :

$\omega_2=0.574\ rad/s$

Part B,

The initial rotational kinetic energy is given by :

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$k_f=\dfrac{1}{2}(I_1+mr^2)\omega_1^2$

$k_f=\dfrac{1}{2}\times (900+70\times (2.9)^2)(0.574)^2$

$k_f=245.24\ rad/s$

Hence, this is the required solution.

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